Phase Margin Question

Discussion in 'Homework Help' started by jegues, Feb 8, 2014.

  1. jegues

    Thread Starter Well-Known Member

    Sep 13, 2010
    735
    43
    For convenience I've attached both the question and my attempt at the solution. Please see figures attached.

    From the quadratic equation for the desired phase margin, I selected the w = 2.682 solution because I thought that we couldn't have negative values for w. (i.e. not physical)

    However the correct solution comes about when w = +1.491, but the solution to my equation indicates w = -1.491.

    How do we know when a negative frequency indicates an invalid solution, or a valid one where we can simply neglect the negative sign?
     
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    I believe it's a simple sign error in your equations:

    You have

    -130^o=-90^o-atan(\frac{\omega}{4-\omega^2})

    which then becomes

    -40^o=-atan(\frac{\omega}{4-\omega^2})

    then

    40^o=atan(\frac{\omega}{4-\omega^2})

    then

    tan(40^o)=(\frac{\omega}{4-\omega^2})

    and so on ....
     
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